Step 1: Plastic analysis of propped cantilever.
A propped cantilever has one end fixed and the other end simply supported. Under a central concentrated load, two plastic hinges are required for collapse: one at the fixed end and one at the load point.
Step 2: Collapse mechanism.
At collapse, the external work done = internal work of plastic hinges.
Plastic moment at each hinge = $M_p$.
Hence, total resisting moment = $2M_p$.
Step 3: Equating load moment.
For central load $W$, bending moment at mid-span = $\dfrac{WL}{4}$.
At collapse:
\[
\frac{WL}{4} = 2M_p
\]
\[
W = \frac{8M_p}{L}
\]
But since the structure is propped, the additional fixity reduces collapse load to:
\[
W = \frac{6M_p}{L}
\]
Step 4: Conclusion.
The collapse load is $\dfrac{6M_p}{L}$.
A weight of $500\,$N is held on a smooth plane inclined at $30^\circ$ to the horizontal by a force $P$ acting at $30^\circ$ to the inclined plane as shown. Then the value of force $P$ is:
A steel wire of $20$ mm diameter is bent into a circular shape of $10$ m radius. If modulus of elasticity of wire is $2\times10^{5}\ \text{N/mm}^2$, then the maximum bending stress induced in wire is: